Working with JSON files in C Sharp (C#)

Introduction

JSON files let programs save and share data in a simple text format. Working with JSON files helps your program read and write data easily.

Saving user settings so they stay the same next time you open the app.

Reading data from a web service that sends information in JSON format.

Storing a list of items like a shopping list or contacts.

Sharing data between different programs or devices.

Loading configuration details when your program starts.

Syntax

C Sharp (C#)

using System.Text.Json;
using System.IO;

// To read JSON from a file
string jsonString = File.ReadAllText("file.json");
var data = JsonSerializer.Deserialize<YourType>(jsonString);

// To write JSON to a file
string output = JsonSerializer.Serialize(data);
File.WriteAllText("file.json", output);

Use JsonSerializer from System.Text.Json namespace for easy JSON handling.

Replace YourType with the class that matches your JSON structure.

Examples

This reads a JSON file into a dictionary where keys and values are strings.

C Sharp (C#)

using System.Text.Json;
using System.IO;

// Reading JSON into a dictionary
string jsonString = File.ReadAllText("data.json");
var dict = JsonSerializer.Deserialize<Dictionary<string, string>>(jsonString);

This saves a simple object with name and age to a JSON file.

C Sharp (C#)

using System.Text.Json;
using System.IO;

// Writing an object to JSON file
var person = new { Name = "Anna", Age = 28 };
string json = JsonSerializer.Serialize(person);
File.WriteAllText("person.json", json);

Sample Program

This program saves a user object to a JSON file and then reads it back to show the data.

C Sharp (C#)

using System;
using System.IO;
using System.Text.Json;

class Program
{
    public class User
    {
        public string Name { get; set; }
        public int Age { get; set; }
    }

    static void Main()
    {
        // Create a user object
        var user = new User { Name = "John", Age = 30 };

        // Convert user to JSON string
        string jsonString = JsonSerializer.Serialize(user);

        // Save JSON string to file
        File.WriteAllText("user.json", jsonString);

        // Read JSON string back from file
        string readJson = File.ReadAllText("user.json");

        // Convert JSON string back to User object
        var userFromFile = JsonSerializer.Deserialize<User>(readJson);

        // Show user info
        Console.WriteLine($"Name: {userFromFile.Name}, Age: {userFromFile.Age}");
    }
}

Important Notes

Make sure your classes have public properties with getters and setters for JSON serialization.

If the JSON file is missing or corrupted, reading it will cause an error, so handle exceptions in real apps.

Summary

JSON files store data in a simple text format easy to read and write.

Use JsonSerializer to convert between objects and JSON strings.

Read JSON from files with File.ReadAllText and write with File.WriteAllText.

Practice

1. What is the main purpose of using JSON files in C# programming?

easy

A. To store and exchange data in a simple text format

B. To compile C# code faster

C. To create graphical user interfaces

D. To manage database connections

Solution

1. Step 1: Understand JSON file usage

   JSON files store data in a readable text format that can be shared or saved easily.

2. Step 2: Identify the correct purpose

   Among the options, only storing and exchanging data matches JSON's role.

3. Final Answer:

   To store and exchange data in a simple text format -> Option A

4. Quick Check:

   JSON = data storage and exchange [OK]

Hint: JSON files hold data as text for easy sharing [OK]

Common Mistakes:

• Thinking JSON compiles code
• Confusing JSON with UI design
• Assuming JSON manages databases

2. Which of the following is the correct way to read a JSON file content into a string in C#?

easy

A. string json = JsonSerializer.ReadFile("data.json");

B. string json = File.ReadAllText("data.json");

C. string json = File.ReadJson("data.json");

D. string json = JsonConvert.Read("data.json");

Solution

1. Step 1: Recall file reading method in C#

   The method File.ReadAllText reads all text from a file into a string.

2. Step 2: Check method correctness

   Only File.ReadAllText("data.json") is valid syntax to read JSON as string.

3. Final Answer:

   string json = File.ReadAllText("data.json"); -> Option B

4. Quick Check:

   File.ReadAllText reads file content [OK]

Hint: Use File.ReadAllText to read JSON file content [OK]

Common Mistakes:

• Using non-existent methods like ReadJson
• Confusing JsonSerializer with file reading
• Using JsonConvert without importing Newtonsoft

3. Given the following code, what will be the output?

using System.Text.Json;

var json = "{\"Name\":\"Alice\", \"Age\":30}";
var person = JsonSerializer.Deserialize<Person>(json);
Console.WriteLine(person.Name + " is " + person.Age + " years old.");

public class Person {
    public string Name { get; set; }
    public int Age { get; set; }
}

medium

A. Alice is 30 years old.

B. Name is Alice, Age is 30

C. System.Text.Json.JsonException

D. null is 0 years old.

Solution

1. Step 1: Deserialize JSON string to Person object

   The JsonSerializer converts the JSON string into a Person object with Name = "Alice" and Age = 30.

2. Step 2: Print the formatted string

   The Console.WriteLine outputs "Alice is 30 years old." by accessing person.Name and person.Age.

3. Final Answer:

   Alice is 30 years old. -> Option A

4. Quick Check:

   Deserialization + property access = output string [OK]

Hint: Deserialize JSON then access properties for output [OK]

Common Mistakes:

• Expecting JSON string printed directly
• Confusing property names or types
• Ignoring deserialization step

4. What is wrong with this code snippet that tries to write an object to a JSON file?

using System.Text.Json;
using System.IO;

var person = new Person { Name = "Bob", Age = 25 };
File.WriteAllText("person.json", person.ToString());

public class Person {
    public string Name { get; set; }
    public int Age { get; set; }
}

medium

A. File.WriteAllText cannot write to JSON files

B. Person class must be static to serialize

C. person.ToString() does not convert the object to JSON format

D. Missing using directive for System.IO

Solution

1. Step 1: Analyze how object is converted to JSON

   Calling person.ToString() returns the class name, not JSON text.

2. Step 2: Identify correct serialization method

   Use JsonSerializer.Serialize(person) to convert the object to JSON string before writing.

3. Final Answer:

   person.ToString() does not convert the object to JSON format -> Option C

4. Quick Check:

   To write JSON, serialize object first [OK]

Hint: Use JsonSerializer.Serialize, not ToString(), to get JSON [OK]

Common Mistakes:

• Using ToString() instead of serialization
• Assuming WriteAllText can't write JSON
• Thinking class must be static to serialize

5. You want to read a JSON file containing a list of products and convert it into a List<Product> in C#. Which code snippet correctly accomplishes this?

public class Product {
    public string Name { get; set; }
    public double Price { get; set; }
}

// JSON file content example: [{"Name":"Pen","Price":1.5},{"Name":"Book","Price":12.99}]

// Which code correctly reads and deserializes the JSON file?

hard

A. var products = JsonSerializer.Deserialize(File.ReadAllText("products.json"));
products = null;

B. var products = JsonSerializer.Deserialize("products.json");

C. var products = JsonSerializer.Deserialize(File.ReadAllLines("products.json"));

D. var json = File.ReadAllText("products.json");
var products = JsonSerializer.Deserialize(json);

Solution

1. Step 1: Read entire JSON file as string

   Use File.ReadAllText to get the JSON content from the file.

2. Step 2: Deserialize JSON string to List<Product>

   Use JsonSerializer.Deserialize(json) to convert JSON text into a list of Product objects.

3. Final Answer:

   var json = File.ReadAllText("products.json"); var products = JsonSerializer.Deserialize(json); -> Option D

4. Quick Check:

   ReadAllText + Deserialize = correct [OK]

Hint: Read file text first, then deserialize to list [OK]

Common Mistakes:

• Passing filename directly to Deserialize
• Using ReadAllLines instead of ReadAllText
• Setting deserialized list to null immediately